在最新版的《C++ Primer》中有這么一句話：若lambda的函數(shù)體包含單一return語(yǔ)句之外的內(nèi)容，且未指定返回類型，則返回void。然后我做了如下測(cè)試：#include?<iostream> #include?<algorithm> #include?<vector> using?namespace?std; int?main(void)?{ ????vector<int>?vi{?1,?-2,?3,?-4,?5,?-6?}; ????/*lambda表達(dá)式返回void？*/ ????/*標(biāo)準(zhǔn)寫(xiě)法：[](int?i)?->?int?{? ????????????????????????????????????if(i?<?0)?return?-i;? ????????????????????????????????????else?return?i;?}?*/ ????transform(vi.begin(),?vi.end(),?vi.begin(),?[](int?i)?{ ????????????????????????????????????????????????????if?(i?<?0)?return?-i; ????????????????????????????????????????????????????else?return?i;?}); ????for?(int?i?:?vi) ????????cout?<<?i?<<?"?"; ????cout?<<?endl; ????system("pause"); ????return?0; }可不管是在Dev-C++中還是在Visual Studio中，它都能正確編譯并且執(zhí)行。如果lambda在這里返回void，就應(yīng)該會(huì)編譯錯(cuò)誤才是??？這是為什么呢？然后我在Visual Studio上又添加了這樣的一段代碼：auto?f?=?[](int?i)?{ ????????????????????if?(i?<?0)?return?-i; ????????????????????else?return?i;?};在visual assistX的幫助下，鼠標(biāo)移動(dòng)到f，它顯示的返回類型也是int。《C++ Primer》寫(xiě)錯(cuò)的可能性不大，比較可能的是編譯器的問(wèn)題。難道是編譯器都不完全遵循C++標(biāo)準(zhǔn)嗎？或者是C++14有什么新規(guī)定嗎？為防止翻譯的誤差，我也特地找了原版來(lái)看，也是如此的：Lambda with function bodies that contain anything other than a single return statement that do not specify a return type return void.??? via 《C++ Primer》 5th Edition, Page 389.However, if we write the seemingly equivalent program using an if statement, our code won't compile:????//error: can't deduce the return type for the lambda.????transform(vi.begin(), vi.end(), vi.begin(), [](int i) { if(i < 0) return -i; else return i; } );via 《C++ Primer》 5th Edition, Page 396.